# This circuit is causing a lot of argument in our shop could someone please evaluate.

#### Csmith260292

Joined Feb 24, 2020
5

#### OBW0549

Joined Mar 2, 2015
3,503
The resistance from A to B is 6 kΩ.

#### Csmith260292

Joined Feb 24, 2020
5
The resistance from A to B is 6 kΩ.
Thanks. I Kept getting answers that would round to 6k also but was told they were wrong

#### Analog Ground

Joined Apr 24, 2019
416
The four resistors on the right half are shorted out. So, do not appear in any calculation. The four resistors on the left half are straightforward. 6K ohms is correct. BTW, there is no rounding required.

(10K || 10K + 10K) || 10K = 6K "||" means "in parallel".

#### Analog Ground

Joined Apr 24, 2019
416
Is this post in the wrong topic? Maybe one under Hardware Design?

#### crutschow

Joined Mar 14, 2008
25,242

#### Csmith260292

Joined Feb 24, 2020
5
The four resistors on the right half are shorted out. So, do not appear in any calculation. The four resistors on the left half are straightforward. 6K ohms is correct. BTW, there is no rounding required.

(10K || 10K + 10K) || 10K = 6K "||" means "in parallel".
Thanks for explaining! And sorry if I put this in the wrong category

#### Csmith260292

Joined Feb 24, 2020
5
The four resistors on the right half are shorted out. So, do not appear in any calculation. The four resistors on the left half are straightforward. 6K ohms is correct. BTW, there is no rounding required.

(10K || 10K + 10K) || 10K = 6K "||" means "in parallel".
Actually with this equation you gave I keep getting 4K so now what am I doing wrong?
because(10K || 10K + 10K)=6.6repeating then 6.6K || 10K= 4K ?

#### Ylli

Joined Nov 13, 2015
976
Take it step by step. The right group is shorted out, so forget about them. In the left group, the diagonal resistor is in parallel with the lower resistor. 10K || 10K = 5K.
That 5K is in series with the top 10K. 10K + 5K = 15K. That 15K is in parallel with the input 10K, so 10K || 15K = 6K.

• sagor

#### Csmith260292

Joined Feb 24, 2020
5
Take it step by step. The right group is shorted out, so forget about them. In the left group, the diagonal resistor is in parallel with the lower resistor. 10K || 10K = 5K.
That 5K is in series with the top 10K. 10K + 5K = 15K. That 15K is in parallel with the input 10K, so 10K || 15K = 6K.

#### Analog Ground

Joined Apr 24, 2019
416
Actually with this equation you gave I keep getting 4K so now what am I doing wrong?
because(10K || 10K + 10K)=6.6repeating then 6.6K || 10K= 4K ?
(10K || 10K) is 5K. Then, 5K + 10K is 15K. Then 15K || 10K is 6K.

R1 || R2 = (R1*R2)/(R1+R2)

It looks like you are calculating 10K || (10K + 5K). The order of operation is (10K || 10K) + 10K.

#### Wolframore

Joined Jan 21, 2019
1,748
Working through all the parallel and series combinations I get 6.33k I worked it out from the back ending with the final parallel pair which is 17.27k ll 10k which is 6.33k

(10k+ 10k) ll 10k
then Bunch of series ... I take it back the back end is shorted. Haha.. it’s 6k